Theorem
Consider a family of closed intervals, $I_1 = [a_1, b_1], I_2 = [a_2, b_2], \ldots$.
If $a_n \leq a_{n+1}$ and $b_{n+1} \leq b_n$ for all $n$ then there is an $x$ which is in every $I_n$, that is, there is an $x \in \displaystyle\bigcap_{n=1}^{\infty} I_n$.

If, however $I_n$ is an open interval, then the Theorem would fail. A counterexample from the book is $\bigg(0, \dfrac{1}{n}\bigg)$ which "kinda" makes sense but honestly I don't fully understand it yet since it is slightly different with what I came up with. My counter example was $\bigg(\dfrac{-1}{n}, \dfrac{1}{n}\bigg)$ which base on the assumption that if $x \in I_n$, then $a_n < x < b_n$ for all $n$. And my reasoning was, if these two sequences $\dfrac{-1}{n}$ and $\dfrac{1}{n}$ meet at the same limit, then there no such $x$ can satisfy $0 < x < 0$. So my question is, is my counterexample correct? Any suggestion or idea would be greatly appreciated.

If $x\ne 0$, however, there is a positive integer $n$ such that $0<\frac1n<|x|$; then either $x<-\frac1n$, or $x>\frac1n$, and in either case $x\notin\left(-\frac1n,\frac1n\right)$. This shows that $0$ is the only real number in the intersection $(1)$, so

$$\bigcap_{n\in\Bbb Z^+}\left(-\frac1n,\frac1n\right)=\{0\}\;.$$

In the case the intervals $\left(0,\frac1n\right)$, you can argue exactly as I just did to show that the intersection is empty: if $x\le 0$, then $x\notin(0,1)$, so certainly $x\notin\bigcap_{n\in\Bbb Z^+}\left(0,\frac1n\right)$, and if $x>0$, choose $n$ big enough so that $0<\frac1n<x$.

Added: You might find it good practice to think about this. Suppose that $\langle a_n:n\in\Bbb N\rangle$ is a strictly increasing sequence, $\langle b_n:n\in\Bbb N\rangle$ is a strictly decreasing sequence, and $a_m<b_m$ for all $m,n\in\Bbb N$:

$$a_0<a_1<a_2<a_3<\ldots\quad\ldots<b_3<b_2<b_1<b_0\;.$$

Then both sequences converge, say to $a$ and $b$, respectively, and $a\le b$. Then

I can't comment on answers, but let me explain why $\cap_{n=1}^\infty (0,\frac{1}{n})$ is empty.
Pick an $n$ and pick $x \in (0,\frac{1}{n})$. Then by the Archimedean property there's an $n_1 \in \mathbb N$ such that $n_1 > 1/x$, or equivalently, $1/n_1 < x$. Then $x \notin (0,\frac{1}{n_1})$. So the intersection of such sets is empty.

You can also phrase this in as a more proper contradiction. If the intersection isn't empty then there's an $x$ such that $x > 0$. But then you find an $n$ so that $1/n < x$, and finish it from there.

The key issue with what you are doing is you are assuming that the intersection of all such sets is open. However, this is not the case. In fact, in Rudin's Principles of Mathematical Analysis, he uses this precise example to show that the infinite intersection of open sets can, in fact, be closed. Since, there is no reason to think that the intersection you describe is open, there is no reason to say that $0<x<0$ but rather you can say that $0\leq x \leq 0$, which, of course implies $x=0 \in \{0\}$, which is the intersection.